对 boost.fusion 向量中元素的引用

发布于 2024-11-16 23:14:18 字数 1080 浏览 4 评论 0原文

编辑 - 请忽略 - 问题是围绕一个简单的拼写错误解决的。我需要休息一下。

如何访问对 boost 融合向量元素的引用？

与 boost.tuples 的 tuples::get(variable)（返回引用）不同，fusion::at_c(variable) 返回一个常量，并且此给我带来困难。

下面说明了我的问题

#include <iostream>
#include <boost/tuple/tuple.hpp> 
#include <boost/fusion/include/vector.hpp>
#include <boost/fusion/include/at.hpp>

using namespace boost;
int
main  (int ac, char **av)
{
   fusion::vector<int, char, std::string> vec(1, 'x', "howdy");
   tuples::tuple <int, char, std::string> tup(1, 'x', "howdy");
   std::cout<< fusion::at_c<0>(vec)<<std::endl;  //outputs 1
   std::cout<< tuples::get<0> (tup) <<std::endl; //outputs 1
   //fusion::at<0>(vec) = 2; //doesn't compile
   tuples::get<0>(tup) = 2;    //works fine
   std::cout<< fusion::at_c<0>(vec) <<std::endl; //can't make this output 2.
   std::cout<< tuples::get<0> (tup) <<std::endl; //outputs 2

}

原文

EDIT - please ignore - the question resolved around a simple typo. I need a break.

How do I access a reference to an element of a boost fusion vector?

Unlike boost.tuples's tuples::get<i>(variable) (returns a reference), the fusion::at_c<i>(variable) returns a constant and this causing me difficulties.

The following illustrates my problem

#include <iostream>
#include <boost/tuple/tuple.hpp> 
#include <boost/fusion/include/vector.hpp>
#include <boost/fusion/include/at.hpp>

using namespace boost;
int
main  (int ac, char **av)
{
   fusion::vector<int, char, std::string> vec(1, 'x', "howdy");
   tuples::tuple <int, char, std::string> tup(1, 'x', "howdy");
   std::cout<< fusion::at_c<0>(vec)<<std::endl;  //outputs 1
   std::cout<< tuples::get<0> (tup) <<std::endl; //outputs 1
   //fusion::at<0>(vec) = 2; //doesn't compile
   tuples::get<0>(tup) = 2;    //works fine
   std::cout<< fusion::at_c<0>(vec) <<std::endl; //can't make this output 2.
   std::cout<< tuples::get<0> (tup) <<std::endl; //outputs 2

}

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